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My attempt:

If $Y=cX+d,$ then $X= (Y-d)/c,$ since $c>0.$

Then $(Y-d)/c \sim \operatorname{Uniform}[L,R],$ since $c>0.$

Isolating for $Y$ gets you $L \le (Y-d)/c \le R,$ since $c>0.$ 

cL <= Y-d <= cR 
cL + d <= Y <= cR+d 
Y ~ Uniform[cL+d,cR+d] 

Y ~ Uniform[(-1)0+1,(-1)1+1] 
Y ~ Uniform[1, 0] 
1 <= -X+1 <= 0 
0 <= -X <= -1 
0 <= X <= 1 
X ~ Uniform[0,1] 

Y = 1-X ~ Uniform[0,1] 
cX + d = 1-X ~ Uniform[0,1] 
X ~ Uniform[0,1] = 1 - (cX + d) 
X ~ Uniform[0,1] = 1- cX - d 
X ~ Uniform[0,1] = 1 + X - 1 
X ~ Uniform[0,1] = X

Is all this correct?

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    This was quite a jumble as posted. I tried to put it into MathJaX, but there are no English words connecting equations and relationships, so I finished by spacing it out and presenting it as text to be more readable. Please re-edit to make it more clear. If you were a student in my class and turned it in for hwk, it wouldn't get much credit without some narrative. – 2017-01-24
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    If $c<0$ and $LcR+d$ hence what is $[cL+d,cR+d]$? – 2017-01-24
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    You need $[cR+d, cL+d]$ rather than $[cL+d, cR+d]. \qquad$ – 2017-01-24
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    My Mistake, it is c>0 – 2017-01-24
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    Then the CDF approach works like a charm, doesn't it? What, you didn't *try* it? – 2017-01-25

1 Answers 1

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I suggest using the moment generator function.

If $X\sim U(L,R)$, then the moment generator function of $X$ i s

$$ M_X(t)=\mathbb{E}[e^{tX}]=\frac{e^{tR}-e^{tL}}{t(R-L)} $$

Then $$ \begin{aligned} M_Y(t)&=\mathbb{E}[e^{tY}]=\mathbb{E}[e^{t(cX+d)}]\\ &=\mathbb{E}[e^{(tc)X}e^{td}]=e^{td}\mathbb{E}[e^{(tc)X}]\\ &=e^{td}\frac{e^{tcR}-e^{tcL}}{tc(R-L)}\\ &=\frac{e^{t(cR+d)}-e^{t(cL+d)}}{t[(cR+d)-(cL+d)]} \end{aligned} $$ Which is the moment generating function of a random variable $U(cL+d,cR+d)$. Therefore, if $X\sim U(L,R)$ then $d+cX=Y\sim U(cL+d,cR+d)$.